If two normals to a parabola `y^2 = 4ax` intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are:

Let A and B be two points on y^(2)=4ax such that normals to the curve at A and B meet at point C, on the curve, then chord AB will always pass through a fixed point whose coordinates, are

If two normals to y^2=4ax are perpendicular to each other, then the chords joining their feet are concurrent at the point :

Chords of the curve 4x^(2) + y^(2)- x + 4y = 0 which substand a right angle at the origin pass thorugh a fixed point whose co-ordinates are :

Two chords are drawn through a fixed point 't' of the parabola y^(2)=4ax and are at right angles. Prove that the join of their other extremities passes through a fixed point.

A normal chord of the parabola y^(2)=4ax subtends a right angle at the vertex if its slope is

If a normal chord at a point on the parabola y^(2)=4ax subtends a right angle at the vertex, then t equals

The normal chord of the parabola y^(2)=4ax subtends a right angle at the focus.Then the end point of the chord is

The normal chord of the parabola y^(2)=4ax subtends a right angle at the vertex.Then the length of chord is